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JEE Main 202628 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
28 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Probability

JEE Main 2026 · 28 January · Evening Shift · Mathematics · Probability

The probability distribution of a random variable X is given below...

The probability distribution of a random variable XX is given below : (X4k307k327k347k367k387k407k6kP(X)2151152151511521515115\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline X & 4k & \dfrac{30}{7}k& \dfrac{32}{7}k & \dfrac{34}{7}k& \dfrac{36}{7}k & \dfrac{38}{7}k& \dfrac{40}{7}k & 6k \\ \hline P(X) & \dfrac{2}{15} & \dfrac{1}{15} & \dfrac{2}{15} & \dfrac{1}{5} & \dfrac{1}{15} & \dfrac{2}{15} & \dfrac{1}{5} & \dfrac{1}{15} \\ \hline \end{array} If E\mathrm{E}(X\mathrm{X})=26315\frac{263}{15}, then P\mathrm{P}(X\mathrm{X}<20)isequalto:is equal to :

Options

  1. A1115\frac{11}{15}
  2. B35\frac{3}{5}
  3. C1415\frac{14}{15}
  4. D815\frac{8}{15}

Correct answer

A. 1115\frac{11}{15}

Step-by-step solution

E(X)=∑XiP(Xi)=526k15×7=26315E(X) = \sum X_i P(X_i) = \dfrac{526k}{15 \times 7} = \dfrac{263}{15}
⇒k=72\Rightarrow k = \dfrac{7}{2}
Substituting k=72k = \dfrac{7}{2}, the values of XX become:
XX: 14, 15, 16, 17, 18, 19, 20, 21
P(X)P(X): 215\dfrac{2}{15}, 115\dfrac{1}{15}, 215\dfrac{2}{15}, 15\dfrac{1}{5}, 115\dfrac{1}{15}, 215\dfrac{2}{15}, 15\dfrac{1}{5}, 115\dfrac{1}{15}
P(X<20)=∑X=1419P(X)P(X < 20) = \sum_{X=14}^{19} P(X)
=215+115+215+15+115+215=1115= \dfrac{2}{15} + \dfrac{1}{15} + \dfrac{2}{15} + \dfrac{1}{5} + \dfrac{1}{15} + \dfrac{2}{15} = \dfrac{11}{15}
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